$\limsup_{n\to\infty} x_n $ and $\liminf_{n\to\infty} x_n $ naturally arise when trying to understand convergent subsequences of the sequence $ (x_n) $. Here's the idea ( We'll be looking at real sequences throughout ) :
Convergent sequences have nice properties ( e.g. boundedness, $ \lim_{n\to\infty} (xa_n + yb_n) = x(\lim_{n\to\infty} a_n) + y(\lim_{n\to\infty} b_n) $, etc ), and the notion is quite central since many other notions can be restated in terms of this ( e.g. "$ f : A (\subseteq \mathbb{R} ) \rightarrow \mathbb{R} $ is continuous at a point $ p \in A $" if and only if "for every sequence $ (a_n) $ in $ A $ converging to $ p $ we have $ f(a_n) $ converging to $ f(p) $" ). In general a sequence doesn't converge, but the next best thing we can have is a convergent subsequence (Recall a subsequence of $ x_1, x_2, x_3, \cdots $ is just a sequence $ x_{j_1}, x_{j_2}, x_{j_3}, \cdots $
with $ j_1 < j_2 < j_3 < \cdots $ ). When does a sequence $ (x_n) $ have a convergent subsequence ? And when it does, what can we say about $ \{ \text{ limits of convergent subsequences of } (x_n) \: \} $ ?
Let's first focus on the second question, i.e. on what happens if a sequence $ (x_n) $ does have a convergent subsequence $ (x_{n_k}) $. Assuming boundedness of $ (x_n) $ we have $ \beta_{n_k} \leq x_{n_k}, x_{n_{k+1}}, x_{n_{k+2}}, \cdots \leq \alpha_{n_k} $ where $ \alpha_j := \sup \{ x_j, x_{j+1}, x_{j+2}, \cdots \} $ and $ \beta_j := \inf \{ x_j, x_{j+1}, x_{j+2}, \cdots \} $. Since $ (\alpha_j) $ is non-increasing & bounded below it has a limit $ \alpha = \inf \alpha_j $, and similarly $ (\beta_j) $ being non-decreasing & bounded above has a limit $ \beta = \sup \beta_j $. Now taking $ k \to \infty $ gives $ \beta \leq \lim_{k\to\infty} x_{n_k} \leq \alpha $.
To summarise : Let $ (x_n) $ be a bounded sequence with a convergent subsequence $ (x_{n_k}) $. Then $ \lim x_{n_k} \in [ \beta, \alpha ] $, where $ \alpha $ and $ \beta $ are the limits of $ \alpha_j := \sup \{ x_j, x_{j+1}, \cdots \} $ and $ \beta_j := \inf \{ x_j, x_{j+1}, \cdots \} $ respectively.
Now let's tackle the first question ( notice in the above discussion we only needed boundedness of $ (x_n) $ to talk about $ \alpha_j, \beta_j $ and their limits $ \alpha $, $ \beta $ ). If $ (x_n) $ is a bounded sequence, because intuitively "there are terms of sequence sticking near $ \alpha_j $s, and the $ \alpha_j $s converge to $ \alpha $" and similarly for $ \beta $, we expect there are subsequences converging to $ \alpha $ & $ \beta $. It's actually true :
Let $ (x_n) $ be a bounded sequence. As usual $ \alpha_j := \sup \{ x_j, x_{j+1}, \cdots \} $, $ \beta_j := \inf \{ x_j, x_{j+1}, \cdots \} $, with respective limits $ \alpha = \inf \alpha_j $ and $ \beta = \sup \beta_j $. There exists an $ n_1 $ such that $ \alpha_1 - 1 < x_{n_1} \leq \alpha_1 $. Now there exists $ n_2 ( \geq n_1 + 1 > n_1 ) $ such that $ \alpha_{{n_1} + 1} - \frac{1}{2} < x_{n_2} \leq \alpha_{{n_1} + 1} $. And now there exists $ n_3 ( \geq n_2 + 1 > n_2 ) $ such that $ \alpha_{{n_2}+1} - \frac{1}{3} < x_{n_3} \leq \alpha_{{n_2}+1} $, and so on. Therefore we get $ n_1 < n_2 < \cdots $ such that $ \alpha_{{n_{j-1}}+1} - \frac{1}{j} < x_{n_{j}} \leq \alpha_{{n_{j-1}}+1} $ for all $ j \geq 1 $ ( we'll take $ n_0 = 0 $, to make sense of the $ j = 1 $ inequality ). So taking $ j \to \infty $, we finally get $ \lim_{j \to \infty} x_{n_j} = \alpha $. Similarly one can construct a subsequence of $ (x_n) $ converging to $ \beta $.
To summarise the entire discussion :
Let $ (x_n) $ be a bounded sequence. Then $ S := \{ \text{ limits of convergent subsequences of } (x_n) \: \} $ is non-empty, with $ \max(S) = \alpha $ and $ \min(S) = \beta $, where $ \alpha = \inf \alpha_j $ and $ \beta = \sup \beta_j $ are the limits of $ \alpha_j := \sup \{ x_j, x_{j+1}, \cdots \} $ and $ \beta_j := \inf \{ x_j, x_{j+1}, \cdots \} $ respectively.
Remark : Here the fact that $ S \neq \varnothing $, i.e. that every bounded sequence of reals has a convergent subsequence, is traditionally called Bolzano-Weierstrass theorem. It is central to Analysis, and can also be proved by a bisection argument.